Math 598
Feb 11, 2005
1
Geometry and Topology II
Spring 2005, PSU
Lecture Notes 6
2.5
The inverse function theorem
Recall that if
f
:
M
→
N
is a diffeomorphism, then
df
p
is nonsingular at all
p
∈
M
(by the chain rule and the observation that
f
◦
f

1
is the identity
function on
M
). The main aim of this section is to prove a converse of this
phenomenon:
Theorem 1
(The Inverse Function Theorem)
.
Let
f
:
M
→
N
be a smooth
map, and
dim(
M
) = dim(
N
)
.
Suppose that
df
p
is nonsingular at some
p
∈
M
.
Then
f
is a
local diffeomorphism
at
p
, i.e., there exists an open
neighborhood
U
of
p
such that
1.
f
is onetoone on
U
.
2.
f
(
U
)
is open in
N
.
3.
f

1
:
f
(
U
)
→
U
is smooth.
In particular,
d
(
f

1
)
f
(
p
)
= (
df
p
)

1
.
A simple fact which is applied a number of times in the proof of the above
theorem is
Lemma 2.
Let
f
:
M
→
N
, and
g
:
N
→
L
be diffeomorphisms, and set
h
:=
g
◦
f
. If any two of the mappings
f
,
g
,
h
are diffeomorphisms, then so
is the third.
In particular, the above lemma implies
Proposition 3.
If Theorem 1 is true in the case of
M
=
R
n
=
N
, then, it
is true in general.
1
Last revised: September 30, 2009
1
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Proof.
Suppose that Theorem 1 is true in the case that
M
=
R
n
=
N
, and
let
f
:
M
→
N
be a smooth map with
df
p
nonsingular at some
p
∈
M
. By
definition, there exist local charts (
U, φ
) of
M
and (
V, ψ
) of
N
, centered at
p
and
f
(
p
) respectively, such that
˜
f
:=
φ

1
◦
f
◦
ψ
is smooth. Since
φ
and
ψ
are diffeomorphisms,
dφ
p
and
dψ
f
(
p
)
are nonsingular.
Consequently, by
the chain rule,
d
˜
f
o
is nonsingular, and is thus a local diffeomorphism. More
explicitly, there exists open neighborhoods
A
and
B
of the origin
o
of
R
n
such that
˜
f
:
A
→
B
is a diffeomorphism.
Since
φ
:
φ

1
(
A
)
→
A
is also a
diffeomorphism, it follows that
φ
◦
˜
f
:
φ

1
(
A
)
→
B
is a diffeomorphism. But
φ
◦
˜
f
=
f
◦
ψ
. So
f
◦
ψ
:
φ

1
(
A
)
→
B
is a diffeomorphism. Finally, since
ψ
:
ψ

1
(
B
)
→
B
is a diffeomorphism, it follows, by the above lemma, that
f
:
φ

1
(
A
)
→
ψ

1
(
B
) is a diffeomorphism.
So it remains to prove Theorem 1 in the case that
M
=
R
n
=
N
.
To
this end we need the following fact. Recall that a metric space is said to be
complete provided that every Cauchy sequence of that space converges.
Lemma 4
(The contraction Lemma)
.
Let
(
X, d
)
be a complete metric space,
and
0
≤
λ <
1
.
Suppose that there exists mapping
f
:
X
→
X
such that
d
(
f
(
x
1
)
,
(
x
2
))
≤
λd
(
x
1
, x
2
)
, for all
x
1
,
x
2
∈
X
. Then there exists a
unique
point
x
∈
X
such that
f
(
x
) =
x
.
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 Spring '08
 Staff
 Geometry, Topology, Chain Rule, Derivative, The Chain Rule, Continuous function, Inverse function, Inverse Function Theorem

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